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Math Help - Commutative Algebra question

  1. #1
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    Commutative Algebra question

    Preliminaries

    f*: Spec(S)->Spec(R)

    Where Spec denotes the set of all prime ideals or R (or S).

    V(I) is a subset of R that contains all the prime ideals that contain I.

    Also, suppose a prime ideal P in S. f^(-1)(P) is a prime ideal in R.

    Prove that f* is continuous.

    I'm not sure if this is helpful. I proved that the preimage of a prime ideal in S is a prime ideal in R. Working off this knowledge, suppose a V(A) in R. Does this imply that there exists a p in P such that f^(-1)(p)=V(A)?
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  2. #2
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    Quote Originally Posted by cmj1988 View Post
    Preliminaries

    f*: Spec(S)->Spec(R)

    Where Spec denotes the set of all prime ideals or R (or S).

    V(I) is a subset of R that contains all the prime ideals that contain I.

    Also, suppose a prime ideal P in S. f^(-1)(P) is a prime ideal in R.

    Prove that f* is continuous.

    I'm not sure if this is helpful. I proved that the preimage of a prime ideal in S is a prime ideal in R. Working off this knowledge, suppose a V(A) in R. Does this imply that there exists a p in P such that f^(-1)(p)=V(A)?
    I think your problem definition is incomplete. So I assume,

    f: R \rightarrow S is a ring homomorphism mapping 1_R to 1_S and V(I) is a Zariski-closed subset of Spec(R). To show that f* is continuous, you need to show that (f^*)^{-1}(V(I)) is a Zariski-closed subset in Spec(S) for an arbitrary Zariski closed subset V(I) in Spec(R). Verify that (f^*)^{-1}(V(I)) is V(f(I)S)=V(I^e) (See here). Thus f* is continuous.

    FYI, you can find some explanations about this problem in Dummit P.734 (proposition 55).
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  3. #3
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    Yes, you completed my definition.

    So I want to show that V(I) in R (which is the collection of prime ideals) gets sent to f*^(-1)(V(I)) such that V(f(I)) in S. Foote's notation is something I'm not use to. In this class our professor just goes off of his notes.

    I'm still a bit shakey on how to prove this. I do understand your explanation though. If we invoke the rule we used in defining f*, we can see that f*^(-1)(V(I)) is f(V(I)) which is contained in some P in S.

    I'm not sure if that works at all.

    Btw, we have not been introduced to this Zariski topology stuff.
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  4. #4
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    Quote Originally Posted by cmj1988 View Post
    Yes, you completed my definition.

    So I want to show that V(I) in R (which is the collection of prime ideals) gets sent to f*^(-1)(V(I)) such that V(f(I)) in S. Foote's notation is something I'm not use to. In this class our professor just goes off of his notes.

    I'm still a bit shakey on how to prove this. I do understand your explanation though. If we invoke the rule we used in defining f*, we can see that f*^(-1)(V(I)) is f(V(I)) which is contained in some P in S.

    I'm not sure if that works at all.

    Btw, we have not been introduced to this Zariski topology stuff.
    (f^*)^{-1}(V(I)) = \bigcap_{x \in I}V((f(x)))=V(\bigcup_{x \in I}(f(x)))=V(I^e)=V(f(I)).

    If you don't topologize the prime spectrums, how do you define a continuous function between them?
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  5. #5
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    In the preface to my sheet, we are to use this definition:

    f: X -> Y is continuous if the inverse image of each closed subset of Y is a closed subset of X.
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